Conditional Probability

Jan 2009
23
0
Hello,

just doing some conditional probability revision and im stuck on the following problems as im just confused about what values to take.

Given that \(\displaystyle P(A) = 0.4, P(B) = 0.6\) and \(\displaystyle P(B|A) = 0.8\)

Find

\(\displaystyle P(B|\overline {A})\) and \(\displaystyle P(\overline {A}|\overline {B})\)

Answers from the previous steps are as follows:
\(\displaystyle P(A \cap B) = 0.32\)
\(\displaystyle P(A \cup B) = 0.68\)
\(\displaystyle P(A|B) = 0.533\)

Cheers
 
Nov 2009
517
130
Big Red, NY
\(\displaystyle P(B) = P(B|A)P(A) + P(B|\bar A)P(\bar A)\)


\(\displaystyle \implies P(B|\bar A) = \frac{P(B) - P(B|A)P(A)}{P(\bar A)}\)


\(\displaystyle P(\bar A) = P(\bar A|B)P(B) + P(\bar A|\bar B)P(\bar B)\)


\(\displaystyle \implies P(\bar A|\bar B) = \frac{P(\bar A) - P(\bar A|B)P(B)}{P(\bar B)}\)
 

Soroban

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May 2006
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Lexington, MA (USA)
Hello, jordanrs!

\(\displaystyle \text{Given: }\;P(A) \,=\,0.4,\;\;P(B) \,=\,0.6,\;\;P(B|A) \,=\, 0.8\)

\(\displaystyle \text{Find: }\;\;(a)\;P\left(B\,|\,\overline {A}\right) \qquad (b)\;P\left(\overline {A}\,|\,\overline {B}\right)\)

Place the information in a chart . . .

. . \(\displaystyle \begin{array}{c||c|c||c|}
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline

P(A) & 0.32 & & 0.40 \\ \hline \\[-4mm]
P\left(\overline{A}\right) & & & \\ \hline \hline
\text{Total} & 0.60 & & 1.00 \\ \hline
\end{array}\)



Fill in the empty cells:

. . \(\displaystyle \begin{array}{c||c|c||c|}
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline

P(A) & 0.32 & 0.08 & 0.40 \\ \hline \\[-4mm]
P\left(\overline{A}\right) & 0.28 & 0.32 & 0.60 \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \\ \hline
\end{array}\)



Answer the questions:

\(\displaystyle (a)\;\;P(B\,|\,\overline{A}) \;=\;\frac{P(B \wedge \overline{A})}{P(\overline{A})} \;=\; \frac{0.28}{0.40} \;=\;0.70 \)

\(\displaystyle (b)\;\;P(\overline{A}\,|\,\overline{B}) \;=\;\frac{P(\overline{A} \wedge \overline{B})}{P(\overline{B})} \;=\; \frac{0.32}{0.40} \;=\;0.80 \)

 
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Reactions: Anonymous1
Nov 2009
517
130
Big Red, NY
Hello, jordanrs!


Place the information in a chart . . .

. . \(\displaystyle \begin{array}{c||c|c||c|}
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline

P(A) & 0.32 & & 0.40 \\ \hline \\[-4mm]
P\left(\overline{A}\right) & & & \\ \hline \hline
\text{Total} & 0.60 & & 1.00 \\ \hline
\end{array}\)



Fill in the empty cells:

. . \(\displaystyle \begin{array}{c||c|c||c|}
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline

P(A) & 0.32 & 0.08 & 0.40 \\ \hline \\[-4mm]
P\left(\overline{A}\right) & 0.28 & 0.32 & 0.60 \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \\ \hline
\end{array}\)



Answer the questions:

\(\displaystyle (a)\;\;P(B\,|\,\overline{A}) \;=\;\frac{P(B \wedge \overline{A})}{P(\overline{A})} \;=\; \frac{0.28}{0.40} \;=\;0.70 \)

\(\displaystyle (b)\;\;P(\overline{A}\,|\,\overline{B}) \;=\;\frac{P(\overline{A} \wedge \overline{B})}{P(\overline{B})} \;=\; \frac{0.32}{0.40} \;=\;0.80 \)

This is a really cool approach.